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Journeyman Electrician practice questions

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Question 1Electrical Theory, Ohm's Law, and Circuit Calculations

A 120 V source is connected across a 24 Ω resistor. What current flows in the circuit?

  1. 0.2 A
  2. 5 A Correct answer.
  3. 120 A
  4. 2.9 A

Explanation

Ohm's law states that current equals voltage divided by resistance, so I = V/R. Substituting gives 120 V ÷ 24 Ω = 5 A. The result is a direct-current value with no other adjustments needed for a purely resistive load.

Question 2Electrical Theory, Ohm's Law, and Circuit Calculations

A resistive load draws 10 A from a 120 V supply. What power does the load consume?

  1. 12 W
  2. 12,000 W
  3. 120 W
  4. 1200 W Correct answer.

Explanation

For a resistive load, power equals voltage multiplied by current, P = V × I. Here 120 V × 10 A = 1200 W. This is equivalent to 1.2 kW.

Question 3Electrical Theory, Ohm's Law, and Circuit Calculations

Resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series. What is the total resistance of the combination?

  1. 600 Ω
  2. 60 Ω Correct answer.
  3. 3.3 Ω
  4. 20 Ω

Explanation

In a series circuit the same current passes through every resistor, so the individual resistances simply add. The total is 10 Ω + 20 Ω + 30 Ω = 60 Ω. Series totals are always larger than the largest individual resistor.

Question 4Electrical Theory, Ohm's Law, and Circuit Calculations

Two 12 Ω resistors are connected in parallel across a supply. What is the equivalent resistance of the pair?

  1. 12 Ω
  2. 24 Ω
  3. 3 Ω
  4. 6 Ω Correct answer.

Explanation

For two resistors in parallel, the reciprocal formula gives 1/RT = 1/12 + 1/12 = 2/12, so RT = 6 Ω. A shortcut for two equal resistors is to divide the value by two. A parallel combination is always smaller than the smallest branch resistance.

Question 5Electrical Theory, Ohm's Law, and Circuit Calculations

Which quantity has the same value at every point in a series circuit?

  1. Voltage
  2. Current Correct answer.
  3. Power
  4. Resistance

Explanation

A series circuit provides only one conductive path, so the same current must pass through every element. Voltage, on the other hand, divides among the loads in proportion to their resistances. Resistance and power both vary from component to component.

Question 6Electrical Theory, Ohm's Law, and Circuit Calculations

A 1500 W space heater operates for 2 hours. How much energy does it use?

  1. 3 kWh Correct answer.
  2. 30 kWh
  3. 1.5 kWh
  4. 0.75 kWh

Explanation

Energy in kilowatt-hours equals power in kilowatts multiplied by time in hours. Converting 1500 W to 1.5 kW and multiplying by 2 hours gives 3 kWh. Utility billing is based on this kilowatt-hour quantity.

Question 7Electrical Theory, Ohm's Law, and Circuit Calculations

A 120 V series circuit has voltage drops of 45 V and 30 V across two resistors. What is the voltage drop across the third resistor?

  1. 195 V
  2. 15 V
  3. 45 V Correct answer.
  4. 75 V

Explanation

Kirchhoff's voltage law requires the sum of the voltage drops around a closed loop to equal the applied voltage. The two known drops total 75 V, so the remaining drop is 120 V − 75 V = 45 V. The full source voltage is thus accounted for.

Question 8Electrical Theory, Ohm's Law, and Circuit Calculations

At a junction, two conductors supply 3 A and 5 A. One conductor leaves the junction carrying 4 A. What current leaves through the remaining conductor?

  1. 2 A
  2. 4 A Correct answer.
  3. 12 A
  4. 8 A

Explanation

Kirchhoff's current law states that total current entering a junction equals total current leaving it. The incoming total is 3 A + 5 A = 8 A, so the unknown outgoing current is 8 A − 4 A = 4 A. Charge is conserved at the node.

Question 9Electrical Theory, Ohm's Law, and Circuit Calculations

A 100 V source is applied across a series combination of 30 Ω and 70 Ω. What voltage appears across the 70 Ω resistor?

  1. 70 V Correct answer.
  2. 100 V
  3. 30 V
  4. 50 V

Explanation

The total resistance is 30 Ω + 70 Ω = 100 Ω, so the current is 100 V ÷ 100 Ω = 1 A. The drop across the 70 Ω resistor is then 1 A × 70 Ω = 70 V. In a series circuit, voltage divides in direct proportion to resistance.

Question 10Electrical Theory, Ohm's Law, and Circuit Calculations

A 24 V source feeds two parallel branches of 8 Ω and 12 Ω. What total current does the source supply?

  1. 5 A Correct answer.
  2. 2 A
  3. 12 A
  4. 3 A

Explanation

Each parallel branch sees the full 24 V, so the branch currents are 24 V ÷ 8 Ω = 3 A and 24 V ÷ 12 Ω = 2 A. Kirchhoff's current law gives a source current of 3 A + 2 A = 5 A. Branch currents add in a parallel circuit.

Question 11Electrical Theory, Ohm's Law, and Circuit Calculations

A conductor is replaced with one of the same material and cross-sectional area but twice the length. How does its resistance change?

  1. It is doubled Correct answer.
  2. It is halved
  3. It is quadrupled
  4. It stays the same

Explanation

Resistance is given by R = ρL/A, so it is directly proportional to length when resistivity and area are fixed. Doubling the length therefore doubles the resistance. Cross-sectional area, by contrast, affects resistance inversely.

Question 12Electrical Theory, Ohm's Law, and Circuit Calculations

A 120 V branch circuit carries 15 A through conductors whose total loop resistance is 0.4 Ω. What is the voltage drop in the conductors?

  1. 48 V
  2. 3 V
  3. 1.6 V
  4. 6 V Correct answer.

Explanation

Voltage drop equals current multiplied by the total conductor resistance of the loop, VD = I × R. Using the round-trip resistance already given, 15 A × 0.4 Ω = 6 V. The load then receives approximately 114 V.

Question 13Electrical Theory, Ohm's Law, and Circuit Calculations

A sinusoidal voltage has a peak value of 170 V. What is its root-mean-square value?

  1. 85 V
  2. 340 V
  3. 170 V
  4. 120 V Correct answer.

Explanation

For a sine wave, RMS value equals peak value divided by the square root of 2, or about 0.707 times the peak. Here 170 V × 0.707 ≈ 120 V. RMS is the value that produces the same heating effect as an equal DC voltage.

Question 14Electrical Theory, Ohm's Law, and Circuit Calculations

What is the inductive reactance of a 0.1 H inductor at 60 Hz?

  1. 0.6 Ω
  2. 377 Ω
  3. 6 Ω
  4. 37.7 Ω Correct answer.

Explanation

Inductive reactance is calculated with XL = 2πfL. Substituting gives 2 × π × 60 Hz × 0.1 H ≈ 37.7 Ω. Reactance rises in direct proportion to both frequency and inductance.

Question 15Electrical Theory, Ohm's Law, and Circuit Calculations

An ideal transformer has a 480 V primary and a 120 V secondary. If the primary current is 5 A, what is the secondary current?

  1. 1.25 A
  2. 5 A
  3. 20 A Correct answer.
  4. 80 A

Explanation

In an ideal transformer, input volt-amperes equal output volt-amperes, so Vp × Ip = Vs × Is. The primary power is 480 V × 5 A = 2400 VA, and 2400 VA ÷ 120 V = 20 A. Stepping voltage down by a factor of four steps current up by the same factor.

Question 16Electrical Theory, Ohm's Law, and Circuit Calculations

Two capacitors of 10 µF and 20 µF are connected in parallel. What is the total capacitance?

  1. 10 µF
  2. 15 µF
  3. 30 µF Correct answer.
  4. 6.7 µF

Explanation

Capacitances in parallel add directly, just as plate area adds, so 10 µF + 20 µF = 30 µF. This is the opposite of the resistor rule for parallel elements. Capacitors in series, however, combine with a reciprocal formula.

Question 17Electrical Theory, Ohm's Law, and Circuit Calculations

Two 20 Ω resistors are connected in parallel, and that pair is connected in series with a 10 Ω resistor. What is the total resistance?

  1. 15 Ω
  2. 50 Ω
  3. 30 Ω
  4. 20 Ω Correct answer.

Explanation

First reduce the parallel pair: 20 Ω in parallel with 20 Ω equals 10 Ω. Then add the series element: 10 Ω + 10 Ω = 20 Ω. Always reduce parallel groups before adding series portions.

Question 18Electrical Theory, Ohm's Law, and Circuit Calculations

A 240 V single-phase load draws 20 A at a power factor of 0.8. What is the true power consumed?

  1. 1920 W
  2. 3840 W Correct answer.
  3. 6000 W
  4. 4800 W

Explanation

True power in an AC circuit is P = V × I × power factor. Substituting gives 240 V × 20 A × 0.8 = 3840 W. The apparent power of 4800 VA includes the reactive component that does no useful work.

Question 19Electrical Theory, Ohm's Law, and Circuit Calculations

A balanced three-phase load operates at 480 V, 20 A, with a power factor of 0.9. What is the approximate real power?

  1. 17.3 kW
  2. 8.6 kW
  3. 15 kW Correct answer.
  4. 30 kW

Explanation

Real power in a balanced three-phase system is P = √3 × V(line) × I(line) × power factor. Computing 1.732 × 480 V × 20 A × 0.9 gives about 14,964 W, or roughly 15 kW. Omitting the √3 factor gives the incorrect single-phase result.

Question 20Electrical Theory, Ohm's Law, and Circuit Calculations

A motor delivers 10 hp of mechanical output at 90 percent efficiency. What input power does it draw?

  1. 9.2 kW
  2. 8.29 kW Correct answer.
  3. 7.46 kW
  4. 6.71 kW

Explanation

First convert the output to watts: 10 hp × 746 W/hp = 7460 W. Because efficiency is output divided by input, the input is 7460 W ÷ 0.90 ≈ 8289 W, or about 8.29 kW. Input power is always greater than output power for a real machine.

Question 21Wiring Methods and Materials

Type NM cable must be secured within what distance of each box or fitting?

  1. 24 inches
  2. 9 inches
  3. 12 inches Correct answer.
  4. 6 inches

Explanation

Section 334.30 requires nonmetallic-sheathed cable to be secured within 12 inches of every box or fitting and at intervals not exceeding 4 1/2 feet. The 12-inch dimension keeps the sheath from being damaged where it enters the enclosure.

Question 22Wiring Methods and Materials

The total bends between pull points in a run of electrical metallic tubing shall not exceed which amount?

  1. 180 degrees
  2. 450 degrees
  3. 360 degrees Correct answer.
  4. 270 degrees

Explanation

Raceway installations limit the sum of all bends between boxes, conduit bodies, or fittings to 360 degrees so conductors can be pulled without excessive tension. Four quarter bends is the maximum regardless of the number of individual bends used.

Question 23Wiring Methods and Materials

Where a raceway contains three or more conductors, the conductors may fill no more than what percentage of the raceway's interior cross-sectional area?

  1. 40 percent Correct answer.
  2. 31 percent
  3. 60 percent
  4. 53 percent

Explanation

Chapter 9, Table 1 limits raceway fill for three or more conductors to 40 percent of the interior cross-sectional area. The smaller percentages apply only to one or two conductors in the raceway.

Question 24Wiring Methods and Materials

Splices and joints in conductors must be enclosed in a box or conduit body that is which of the following?

  1. Listed for wet locations
  2. Concealed in the building finish
  3. Permanently sealed
  4. Readily accessible Correct answer.

Explanation

Section 300.15 requires spliced conductors to be enclosed, and the enclosure must remain accessible so the connection can be inspected or serviced. Concealing a splice inside a finished wall or ceiling is prohibited.

Question 25Wiring Methods and Materials

Where NM cable passes through a bored hole less than 1 1/4 inches from the nearest edge of a framing member, protection must be provided by a steel plate at least how thick?

  1. 1/32 inch
  2. 3/64 inch
  3. 1/16 inch Correct answer.
  4. 1/8 inch

Explanation

Section 300.4(D) requires a steel plate not less than 1/16 inch thick where the cable is closer than 1 1/4 inches to the edge of the framing member. The plate prevents nails and screws driven into the framing from penetrating the cable.

Question 26Wiring Methods and Materials

Type NM cable is permitted to be installed in which of the following locations?

  1. Embedded in poured concrete
  2. In a hoistway
  3. In a wet location
  4. Inside a dry, heated crawlspace Correct answer.

Explanation

Section 334.10 permits NM cable in normally dry locations such as a dry, heated crawlspace, which is why the cable's outer jacket is rated for dry use. Wet locations, hoistways, and embedment in concrete are all prohibited by Section 334.12.

Question 27Wiring Methods and Materials

Electrical metallic tubing must be securely fastened within what distance of each box or fitting?

  1. 3 feet Correct answer.
  2. 10 feet
  3. 6 feet
  4. 5 feet

Explanation

Section 358.30 requires EMT to be secured within 3 feet of each box, conduit body, or fitting, and at intervals not exceeding 10 feet thereafter. The 3-foot rule limits strain on the enclosure entry.

Question 28Wiring Methods and Materials

At each outlet, switch, or junction box, the minimum length of free conductor that must be left for splices or terminations is which of the following?

  1. 4 inches
  2. 2 inches
  3. 6 inches Correct answer.
  4. 3 inches

Explanation

Section 300.14 requires at least 6 inches of free conductor measured from the point where the conductor emerges from its raceway or cable sheath. At least 3 inches must extend outside the box opening.

Question 29Wiring Methods and Materials

A box contains four 12 AWG ungrounded and grounded conductors, one cable clamp, one equipment grounding conductor, and one device yoke. Using 2.25 cubic inches per 12 AWG conductor, what is the minimum box volume required?

  1. 18.00 cubic inches Correct answer.
  2. 22.50 cubic inches
  3. 15.75 cubic inches
  4. 20.25 cubic inches

Explanation

Section 314.16(B) counts the four conductors, one for the clamp, one for the equipment grounding conductor, and two for the device yoke, for a total of eight conductor volumes. Eight times 2.25 cubic inches equals 18.00 cubic inches.

Question 30Wiring Methods and Materials

Schedule 40 rigid PVC conduit of 1/2-inch trade size installed in a straight horizontal run must be supported at intervals not exceeding which distance?

  1. 3 feet Correct answer.
  2. 2 feet
  3. 5 feet
  4. 18 inches

Explanation

Table 352.30 sets the support interval for 1/2-inch through 1-inch PVC at 3 feet. Larger sizes are permitted longer intervals, up to 6 feet for the largest trade sizes.

Question 31Wiring Methods and Materials

Type UF cable is direct buried as a feeder beneath a residential lawn. What is the minimum cover required?

  1. 30 inches
  2. 12 inches
  3. 24 inches Correct answer.
  4. 18 inches

Explanation

Table 300.5 lists 24 inches as the minimum cover for direct-buried UF cable for feeders and general branch circuits. The reduced 12-inch cover applies only to residential branch circuits limited to 120 volts and 20 amperes with ground-fault protection.

Question 32Wiring Methods and Materials

What is the minimum radius to the inner edge of a bend for 1/2-inch electrical metallic tubing?

  1. 4 inches Correct answer.
  2. 3 inches
  3. 2 1/2 inches
  4. 2 inches

Explanation

Table 358.24 specifies a minimum bend radius of 4 inches for 1/2-inch EMT. Bending tighter than the table value can flatten or kink the tubing and damage conductors during a pull.

Question 33Wiring Methods and Materials

Type AC cable must be supported at intervals not exceeding which distance?

  1. 4 1/2 feet Correct answer.
  2. 6 feet
  3. 3 feet
  4. 4 feet

Explanation

Section 320.30 requires armored cable to be secured at intervals not exceeding 4 1/2 feet and within 12 inches of each box or fitting. The same 4 1/2-foot interval applies to NM cable.

Question 34Wiring Methods and Materials

Interlocked-armor Type MC cable must be secured at intervals not exceeding which distance?

  1. 6 feet Correct answer.
  2. 10 feet
  3. 4 1/2 feet
  4. 3 feet

Explanation

Section 330.30 requires Type MC cable to be supported at intervals not exceeding 6 feet and within 12 inches of boxes and fittings, with limited exceptions for certain industrial and cable-tray installations.

Question 35Wiring Methods and Materials

The sum of the cross-sectional areas of all conductors installed in a metal wireway may not exceed what percentage of the wireway's interior cross-sectional area at any point?

  1. 10 percent
  2. 20 percent Correct answer.
  3. 40 percent
  4. 25 percent

Explanation

Section 376.22 limits conductor fill in wireways to 20 percent of the interior cross-sectional area at any cross section. The generous space allows conductors to be laid in without damage and helps dissipate heat.

Question 36Wiring Methods and Materials

Liquidtight flexible metal conduit of 1/2-inch trade size must be supported at intervals not exceeding which distance?

  1. 5 feet
  2. 4 feet
  3. 3 feet Correct answer.
  4. 18 inches

Explanation

Table 350.30 sets a support interval of 3 feet for 1/2-inch through 1-inch liquidtight flexible metal conduit. Larger trade sizes are permitted intervals of 3 1/2 to 6 feet.

Question 37Wiring Methods and Materials

A straight run of rigid PVC conduit is installed between two fixed enclosures and will be exposed to a seasonal temperature swing. What must the installation include?

  1. A bonding jumper
  2. An expansion fitting Correct answer.
  3. A seal-off fitting
  4. An equipment grounding conductor

Explanation

Section 352.44 requires expansion fittings where a PVC raceway is subject to thermal expansion and contraction, such as a straight run between fixed points. Without them, the conduit can bow or pull apart at joints.

Question 38Wiring Methods and Materials

A box used to support a ceiling-suspended paddle fan must be which of the following?

  1. Made of metal
  2. Listed and marked for the application Correct answer.
  3. At least 4 inches deep
  4. Installed with a box extension

Explanation

Section 314.27(C) requires boxes that support ceiling-suspended paddle fans to be listed and marked for that use. The marking confirms the box and its mounting means can carry the fan's weight and vibration.

Question 39Wiring Methods and Materials

How much volume must be counted for each 14 AWG conductor when calculating box fill?

  1. 2.50 cubic inches
  2. 2.00 cubic inches Correct answer.
  3. 1.50 cubic inches
  4. 2.25 cubic inches

Explanation

Table 314.16(B) assigns 2.00 cubic inches to each 14 AWG conductor, 2.25 cubic inches to each 12 AWG conductor, and 2.50 cubic inches to each 10 AWG conductor. The same values apply to clamps, support fittings, and equipment grounding conductors as counted in Section 314.16(B).

Question 40Wiring Methods and Materials

Rigid PVC conduit is direct buried beneath a residential lawn to supply a detached garage. What is the minimum cover required?

  1. 24 inches
  2. 18 inches Correct answer.
  3. 12 inches
  4. 6 inches

Explanation

Table 300.5 requires a minimum cover of 18 inches for rigid PVC conduit installed direct buried in earth. The greater depths in the table apply to roadways, airport runways, and similar areas.

Question 41Wiring Methods and Materials

In a Class I, Division 1 location, a seal-off fitting must be installed within what distance of each conduit entering an enclosure?

  1. 24 inches
  2. 12 inches
  3. 18 inches Correct answer.
  4. 36 inches

Explanation

Section 501.15(A)(1) requires a seal-off fitting in each conduit run entering an enclosure in a Class I, Division 1 location, located within 18 inches of the enclosure. The seal limits the passage of gases or vapors through the conduit system.

Question 42Wiring Methods and Materials

Flexible metal conduit may serve as an equipment grounding conductor where the circuit is protected at 20 amperes or less, the length does not exceed 6 feet, and which other condition is met?

  1. The entire run is installed in a dry location
  2. The conduit is 1 1/4 inches or larger in trade size
  3. Listed grounding fittings are used Correct answer.
  4. The circuit contains no splices or joints

Explanation

Section 250.118(5) recognizes flexible metal conduit as an equipment grounding conductor only for short lengths (6 feet or less) on circuits protected at 20 amperes or less, and only when terminated with fittings listed for grounding. The listed fittings ensure a reliable low-impedance path.

Question 43Wiring Methods and Materials

What is the maximum number of 12 AWG THHN copper conductors permitted in a 1/2-inch electrical metallic tubing raceway?

  1. 6
  2. 12
  3. 14
  4. 9 Correct answer.

Explanation

The 40 percent fill area for 1/2-inch EMT is about 0.122 square inch, and each 12 AWG THHN conductor occupies about 0.0133 square inch. Dividing gives just over nine conductors, so nine is the maximum permitted.

Question 44Wiring Methods and Materials

How much volume must be counted for each 6 AWG conductor in a box-fill calculation?

  1. 4.50 cubic inches
  2. 3.00 cubic inches
  3. 4.00 cubic inches
  4. 5.00 cubic inches Correct answer.

Explanation

Table 314.16(B) assigns 5.00 cubic inches to each 6 AWG conductor. The table values increase with conductor size, and the same volume applies to an equipment grounding conductor of that size.

Question 45Wiring Methods and Materials

A conduit nipple 24 inches or less in length, installed between boxes or enclosures, may be filled to what percentage of its cross-sectional area?

  1. 60 percent Correct answer.
  2. 40 percent
  3. 31 percent
  4. 50 percent

Explanation

Note 4 to Chapter 9, Table 1 permits nipple raceways 24 inches or less in length to be filled to 60 percent of the interior cross-sectional area. The short length makes conductor pulling practical despite the higher fill.

Question 46Wiring Methods and Materials

Type NM cable installed along the side of wood framing must be supported at intervals not exceeding what distance, and within what distance of each box?

  1. 4 1/2 ft, and within 12 in of each box Correct answer.
  2. 3 ft, and within 6 in of each box
  3. 6 ft, and within 8 in of each box
  4. 10 ft, and within 18 in of each box

Explanation

Nonmetallic-sheathed cable must be secured by staples, straps, or similar fittings so that it is supported at intervals not exceeding 4 1/2 ft and within 12 in of every box or fitting. The 12-in rule keeps strain off terminations at the enclosure.

Question 47Wiring Methods and Materials

Where is it permitted to make a splice or joint in conductors that are part of a raceway wiring method?

  1. Anywhere, provided the splice is insulated with tape and compound
  2. In a box, conduit body, or fitting identified for the purpose Correct answer.
  3. Anywhere in the run that remains accessible after construction
  4. At any point in the run where the splice is made with listed connectors

Explanation

Conductor splices and joints must be enclosed in an approved box, conduit body, or fitting designed for the purpose so the connection is protected and accessible. Making up joints inside a raceway or in an open run is not permitted regardless of the connector type.

Question 48Wiring Methods and Materials

A single receptacle is mounted in a box and connected to 12 AWG circuit conductors. How many conductor volumes must be counted for the device when calculating box fill?

  1. One conductor volume, counted at 12 AWG
  2. Two conductor volumes, counted at 12 AWG Correct answer.
  3. Two conductor volumes, counted at 14 AWG
  4. One conductor volume, counted at 18 AWG

Explanation

Each device or equipment, such as a receptacle or switch, is counted as two conductors based on the largest conductor connected to it. Here the largest conductor is 12 AWG, so the receptacle adds the equivalent of two 12 AWG volumes to the box fill.

Question 49Wiring Methods and Materials

Rigid metal conduit of 1-inch trade size installed in a straight horizontal run must be supported at intervals not exceeding which distance?

  1. 14 ft
  2. 10 ft
  3. 16 ft
  4. 12 ft Correct answer.

Explanation

Support spacing for rigid metal conduit increases with trade size; 1-inch raceway is permitted to be supported at intervals not exceeding 12 ft. The run must also be secured within 3 ft of each box or termination fitting.

Question 50Wiring Methods and Materials

Which of the following wiring methods is permitted to be installed in an air-handling plenum used to move environmental air?

  1. Type NM cable
  2. Schedule 40 rigid PVC conduit
  3. Electrical metallic tubing Correct answer.
  4. Electrical nonmetallic tubing

Explanation

In spaces used to move environmental air, only wiring methods specifically permitted for such spaces may be used; metal raceways such as electrical metallic tubing qualify. Nonmetallic methods like PVC conduit, ENT, and NM cable are prohibited because of smoke and flame-spread characteristics of their materials.